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# How to calculate voltage regulation of distribution line

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## Voltage regulation How to calculate voltage regulation of distribution line (on photo: Distribution Lines – Oaxaca, Mexico, 2013 via FlickR)

## Introduction to voltage regulation

Voltage (load) regulation is to maintain a fixed voltage under different load.Voltage regulation is limiting factor to decide the size of either conductor or type of insulation.

In circuit current need to be lower than this in order to keep the voltage drop within permissible values. The high voltage circuit should be carried as far as possible so that the secondary circuit have small voltage drop.

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### Voltage Regulation for 11KV, 22KV, 33KV Overhead Line

% Voltage Regulation = (1.06 x P x L x PF) / (LDF x RC x DF)

Where:

P – Total Power in KVA
L –  Total Length of Line from Power Sending to Power Receiving in KM.
PF – Power Factor in p.u
RC – Regulation Constant (KVA-KM) per 1% drop.

RC = (KV x KV x 10) / ( RCosΦ + XSinΦ)

LDF – Load Distribution Factor.
LDF = 2 for uniformly distributed Load on Feeder.
LDF > 2 If Load is skewed toward the Power Transformer.
LDF = 1 To 2 If Load is skewed toward the Tail end of Feeder.

DF – Diversity Factor in p.u

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### Permissible Voltage Regulation (As per REC)

 Maximum  Voltage Regulation at any Point of Distribution Line Part of Distribution System Urban Area (%) Suburban Area (%) Rural Area (%) Up to Transformer 2.5 2.5 2.5 Up to Secondary  Main 3 2 0.0 Up to Service Drop 0.5 0.5 0.5 Total 6.0 5.0 3.0

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### Voltage Regulation Values

The voltage variations in 33 kV and 11kV feeders should not exceed the following limits at the farthest end under peak load conditions and normal system operation regime.

• Above 33kV (-) 12.5% to (+) 10%.
• Up to 33kV (-) 9.0% to (+) 6.0%.
• Low voltage (-) 6.0% to (+) 6.0%

In case it is difficult to achieve the desired voltage especially in Rural areas, then 11/0.433 kV distribution transformers(in place of normal 11/0.4 kV DT’s) may be used in these areas.

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### Required Size of Capacitor

Size of capacitor for improvement of the Power Factor from Cos ø1 to Cos ø2 is:

Required size of Capacitor (Kvar) = KVA1 (Sin ø1 – [Cos ø1 / Cos ø2] x Sin ø2)

Where KVA1 is Original KVA.

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### Optimum location of capacitors

L = [1 – (KVARC / 2 KVARL) x (2n – 1)]

Where:

L – distance in per unit along the line from sub-station.
KVARC – Size of capacitor bank
n – relative position of capacitor bank along the feeder from sub-station if the total capacitance is to be divided into more than one Bank along the line. If all capacitance is put in one Bank than values of n=1.

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### Voltage Rise due to Capacitor installation:

% Voltage Rise = (KVAR(Cap) x Lx X) / 10 x Vx2

Where:

KVAR (Cap) – Capacitor KVAR
X – Reactance per phase
L – Length of Line (mile)
V – Phase to phase voltage in kilovolts

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## Calculate % Voltage Regulation of Distribution Line

Calculate Voltage drop and % Voltage Regulation at Trail end of following 11 KV Distribution system:

• System have ACSR DOG Conductor (AI 6/4.72, GI7/1.57)
• Current Capacity of ACSR Conductor = 205Amp,
• Resistance = 0.2792Ω and Reactance = 0 Ω,

Permissible limit of % Voltage Regulation at Trail end is 5%. ### Method-1 (Distance Base)

Voltage Drop  = ( (√3x(RCosΦ+XSinΦ)x I ) / (No of Conductor/Phase x1000))x Length of Line

#### Voltage drop at Load A

• Load Current at Point A (I) = KW / 1.732xVoltxP.F
• Load Current at Point A (I) =1500 / 1.732x11000x0.8 = 98 Amp.
• Required No of conductor / Phase =98 / 205 =0.47 Amp =1 No
• Voltage Drop at Point A = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line
• Voltage Drop at Point A =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x1500) = 57 Volt
• Receiving end Voltage at Point A = Sending end Volt-Voltage Drop= (1100-57) = 10943 Volt.
• % Voltage Regulation at Point A = ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
• % Voltage Regulation at Point A = ((11000-10943) / 10943 )x100 = 0.52%
• % Voltage Regulation at Point A =0.52 %

#### Voltage drop at Load B

• Load Current at Point B (I) = KW / 1.732xVoltxP.F
• Load Current at Point B (I) =1800 / 1.732x11000x0.8 = 118 Amp.
• Distance from source= 1500+1800=3300 Meter.
• Voltage Drop at Point B = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line
• Voltage Drop at Point B =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x3300) = 266 Volt
• Receiving end Voltage at Point B = Sending end Volt-Voltage Drop= (1100-266) = 10734 Volt.
• % Voltage Regulation at Point B= ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
• % Voltage Regulation at Point B= ((11000-10734) / 10734 )x100 = 2.48%
• % Voltage Regulation at Point B =2.48 %

#### Voltage drop at Load C

• Load Current at Point C (I) = KW / 1.732xVoltxP.F
• Load Current at Point C (I) =2000 / 1.732x11000x0.8 = 131 Amp
• Distance from source= 1500+1800+2000=5300 Meter.
• Voltage Drop at Point C = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line
• Voltage Drop at Point C =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x5300) = 269 Volt
• Receiving end Voltage at Point C = Sending end Volt-Voltage Drop= (1100-269) = 10731 Volt.
• % Voltage Regulation at Point C= ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
• % Voltage Regulation at Point C= ((11000-10731) / 10731 )x100 = 2.51%
• % Voltage Regulation at Point C =2.51 %

Here Trail end Point % Voltage Regulation is 2.51% which is in permissible limit.

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### Method-2 (Load Base)

% Voltage Regulation =(I x (RcosǾ+XsinǾ)x Length ) / No of Cond.per Phase xV (P-N))x100

#### Voltage drop at Load A

• Load Current at Point A (I) = KW / 1.732xVoltxP.F
• Load Current at Point A (I) =1500 / 1.732x11000x0.8 = 98 Amp.
• Distance from source= 1.500 Km.
• Required No of conductor / Phase =98 / 205 =0.47 Amp =1 No
• Voltage Drop at Point A = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100
• Voltage Drop at Point A =((98x(0.272×0.8+0×0.6)x1.5) / 1×6351) = 0.52%
• % Voltage Regulation at Point A =0.52 %

#### Voltage drop at Load B

• Load Current at Point B (I) = KW / 1.732xVoltxP.F
• Load Current at Point B (I) =1800 / 1.732x11000x0.8 = 118 Amp.
• Distance from source= 1500+1800=3.3Km.
• Required No of conductor / Phase =118 / 205 =0.57 Amp =1 No
• Voltage Drop at Point B = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100
• Voltage Drop at Point B =((118x(0.272×0.8+0×0.6)x3.3)/1×6351) = 1.36%
• % Voltage Regulation at Point A =1.36 %

#### Voltage drop at Load C

• Load Current at Point C (I) = KW / 1.732xVoltxP.F
• Load Current at Point C (I) =2000 / 1.732x11000x0.8 = 131Amp.
• Distance from source= 1500+1800+2000=5.3Km.
• Required No of conductor / Phase =131/205 =0.64 Amp =1 No
• Voltage Drop at Point C = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100
• Voltage Drop at Point C =((131x(0.272×0.8+0×0.6)x5.3)/1×6351) = 2.44%
• % Voltage Regulation at Point A =2.44 %

Here Trail end Point % Voltage Regulation is 2.44% which is in permissible limit.

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Premium membership gives you an access to specialized technical articles and extra premium content (electrical guides and software). ### Jignesh Parmar

Jignesh Parmar has completed M.Tech (Power System Control), B.E (Electrical). He is member of Institution of Engineers (MIE), India. He has more than 13 years experience in transmission & distribution-energy theft detection and maintenance electrical projects.

1. Ghazanfar Ali Khan
Aug 01, 2017

How 3rd harmonics are mitigated in tertiary winding, delta connected.
The answer is pending since 2015. Why no body has turned back with a comprehensive reply,including Jignesh Parmar.

2. NP Singh
Feb 08, 2017

One 11 KV HT line feeder is required to provide power supply to the sub divided 85 Nos. industrial plots. Though these plots are not yet sold but considering the futuristic requirement of the power supply of the individual plot holder, an electrification scheme is required to be passed from DHBVN as it is technically not feasible to get power supply from the DHBVN. The total minimum load as per the norms of the dhbvn is 3915 kv, hence we need to apply for the sanctioning of the load. Could anyone suggest, how the demand factor / diversification factor is taken and what would be the load required for applying.
As per my calculation the load is : 3910 divided by 0.9 = 4344 kva
Demand factor would = 3910 x .625 = 2443.75 kva or 2.45 mva approx.
Request confirm whether the demand factor 0.625 will be divided or multiplied to get the total load?

3. Engr. Ghazanfar Ali Khan
Oct 01, 2016

How third harmonics are mitigated by tertiary winding?

4. Narendrasinh Desai
Aug 05, 2016

sir
i am working in Electricity Department Silvassa. recently i have made calculation as per above method which is as under . kindly guide me is it true?
% Voltage Regulation of Distribution Line
A
66/11KV Athal s/s end Load 250KVA=225KW at PF.-09

Calculate Voltage drop and % Voltage Regulation at Trail end of following 11 KV Distribution system:
• System have ACSR DOG Conductor (AI 6/4.72, GI7/1.57) ATHAL FEEDER II
• Current Capacity of ACSR Conductor = 300Amp,
• Resistance = 0.2792Ω and Reactance = 0 Ω,
Permissible limit of % Voltage Regulation at Trail end is 5%.
Load at point A=250KVA
Taking P.F-0.9 of system factor
Load =250 X 0.9=225KW
Length of Load Point from Athal S/s=1.5KM=1500Mtr
Method- (Distance Base)
Voltage Drop = ( (√3x(RCosΦ+XSinΦ)x I ) / (No of Conductor/Phase x1000))x Length of Line
Voltage drop at Load A
• Load Current at Point A (I) = KW / 1.732xVoltxP.F
• Load Current at Point A (I) =225 / 1.732x11000x0.99 = 13.12 Amp.
• Required No of conductor / Phase =13.12 / 300 =0.044 Amp =1 No
• Voltage Drop at Point A = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line
• Voltage Drop at Point A =((1.732x (0.272×0.99+0×0.141)x13.12) / 1×1000)x1500) = 9.21 Volt
• Receiving end Voltage at Point A = Sending end Volt-Voltage Drop= (11000-9.21) = 10990.79 Volt.
• % Voltage Regulation at Point A = ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
• % Voltage Regulation at Point A = ((11000-10990.79) / 10990.79 )x100 = 0.084%
• % Voltage Regulation at Point A =0.084 %

5. ssrao
Jul 02, 2016

sir we facing probem of 33kv line fault at mtabu clouds before monsoon flesh over the linearc devlop betwwen conductor gi pin of pin insulator

6. Mohamed A Farah
May 16, 2016

What is the longest distance an 11KV line can cover before you upgrade to 33KV line.

We are using 11KV line for the inner city. We are trying to move our generation machinery to a site that is 7 kms away from the city. Some of our staff are saying we should upgrade to a 33KV line for the distance. Is that acceptable. Or should we stick to the 11KV line we use for our inner city distribution.

In other words, does it mean a higher resistance for the 11KV line to justify the 33KV line.

Thanks;

Mohamed A Farah
Sompower Co.

7. Jan 19, 2016

how can i calculate 22KV cable voltage drop ??

Load Current at Point B (I) = KW / 1.732xVoltxP.F
Load Current at Point B (I) =1800 / 1.732x11000x0.8 = 118 Amp.
Distance from source= 1500+1800=3.3Km.
Required No of conductor / Phase =118 / 205 =0.57 Amp =1 No
Voltage Drop at Point B = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100
Voltage Drop at Point B =((118x(0.272×0.8+0×0.6)x3.3)/1×6351) = 1.36%
% Voltage Regulation at Point A =1.36 %

8. danial fazilatpoor
Jan 12, 2016

Dear sir
I have question . Why we choose 5 percent for voltage regulation , I mean i khow there is a standard but the question is what is the idea behind this standard.
thanks a lot.

9. Girish
Oct 27, 2015

Dear Sir,

Could you please share the method/formulae for the voltage drop calculation for overhead LT line feeders.

10. SK Farid
May 05, 2015

How its possible Voltage drop proper calculation?

11. Sushil Kumar
Apr 14, 2015

Dear Sir, i want to calculate the voltage drop of an overhead line using DOG Conductor of length around 3000 meter and load around 400 KW.
When i m calculating the voltage drop per line then it is over permissible (5%).
Total load of Line = 400 KW
Total length of line = 3000 meter
Transformer capacity = 630 KVA.
Pls guide me for make it under permissible.

12. Lombard
Nov 08, 2014

I would like to pull power from a 33kv line to my camp 17km away. I am looking at a installing a 200Kva Transformer, 3 phase, 380V
1/ What size CROSS SECTION mm of AAC Aluminium Overhead cable would I require?
2/ Would I be able to connect other transformers upto 400Kva, in Total?

13. Pulin Panchal
May 01, 2014

Can you give calculation of Voltage Drop of LV system having 2 or 3 run Parallel Power Cable connection ?

• Pankaj Chavan
Jun 25, 2015

Sir,
LT CABLE VOL.DROP FORMULA IS AS BELOW.

% V.D = 1.732 × LOAD CURRENT × LENGTH × ( CABLE RESISTANCE × COS PHI + CABLE REACTANCE × SIN PHI ) ×100 ÷ NO OF RUNS × 415 × 1000

If any other queries in above formula contact me on above mail.

14. mohamed ASIM
Feb 08, 2014

Dear sir,

In calculating voltage drop at the point B, the load current is 118 Amps,(in your example).But while calculating the voltage drop at the point B,you have taken the current value of point A( 98Amps). is that correct..can you explain…

15. MANISH
Nov 02, 2013

what is maximum loading in 100 KVA and 63 KVA DTR in HP?

• Edvard
Nov 02, 2013

At what voltage level?

16. joboyouk
Oct 16, 2013

I just presently left my banking job to return to field of engineering. Your work on this site has been a tremendous refresher for me and even much more. I just want to say thank you for being part of those helping my confidence of finding a place in the field.
However, there’s still much for me to cover just keep up the good work.

jobo3889@gmail.com

17. amendozc
Oct 07, 2013

Buenos días Dr Jignesh Parmar

Es usted muy gentil con su publicación. Sin embargo, agradezco por favor resolverme las siguientes inquietudes:

1. Indicarme la razón por la cual, en los ejemplos de regulación, no se acumuló en la carga A, la carga B y C, teniendo en cuenta que se debería acumular el momento eléctrico al momento de cálculo.

2. Indicarme por favor bibliografía adicional (normatividad) para este tipo de cálculo.

Gracias Dr

18. gaurav rana
Aug 27, 2013

dear sir!
with due to respect i want to say that you done a fantastic job on this portal and its very helpful to all engineers,technician and all people who related to electrical engineering field.
at last my one request to you that give me more info about the interconnecting transformer ICT using auto transformer, please!
rest is fine and thanking you

GAURAV SINGH RANA
gsrana89025@gmail.com

19. TSRG
Jun 11, 2013

Dear Sir,
It is more useful datas.
Many more thanks & grateful to you.

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